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    <meta name="description" content="时间复杂度一个算法需要执行的次数我们记为T(n)，其中n为算法的规模。现在引入某个辅助函数f(n)，当n趋近于无穷大时，T(n)&#x2F;f(n) &#x3D; C (C ≠ 0)。则f(n)和T(n)是同量级函数，记为T(n) &#x3D; O(f(n))，我们称这个为时间复杂度。 ##每种时间复杂度表示的意思 T(n) &#x3D; O(1)123- (void)aFunction0:(in">
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          时间复杂度
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        <h2 id="时间复杂度"><a href="#时间复杂度" class="headerlink" title="时间复杂度"></a>时间复杂度</h2><p>一个算法需要执行的次数我们记为T(n)，其中n为算法的规模。现在引入某个辅助函数f(n)，当n趋近于无穷大时，T(n)&#x2F;f(n) &#x3D; C (C ≠ 0)。则f(n)和T(n)是同量级函数，记为T(n) &#x3D; O(f(n))，我们称这个为时间复杂度。</p>
<p>##每种时间复杂度表示的意思</p>
<h3 id="T-n-x3D-O-1"><a href="#T-n-x3D-O-1" class="headerlink" title="T(n) &#x3D; O(1)"></a>T(n) &#x3D; O(1)</h3><figure class="highlight objc"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">- (<span class="type">void</span>)aFunction0:(<span class="type">int</span>)n &#123;</span><br><span class="line">    <span class="built_in">NSLog</span>(<span class="string">@&quot;%zd&quot;</span>,n); <span class="comment">// 执行 1 次</span></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>不管输入的n是多少，执行次数都是常数。执行次数和输入n值没有任何关系。<br>T(n) &#x3D; 1 &#x3D; O(1)</p>
</blockquote>
<h3 id="T-n-x3D-O-n"><a href="#T-n-x3D-O-n" class="headerlink" title="T(n) &#x3D; O(n)"></a>T(n) &#x3D; O(n)</h3><figure class="highlight objc"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">- (<span class="type">void</span>)aFunction1:(<span class="type">int</span>)n &#123;</span><br><span class="line">    <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++) &#123; <span class="comment">// n + 1 次</span></span><br><span class="line">        <span class="built_in">NSLog</span>(<span class="string">@&quot;%zd&quot;</span>,i); <span class="comment">// n 次</span></span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>执行次数和输入的n值成线性关系。<br>T(n) &#x3D; n + 1 + n &#x3D; 2n + 1 &#x3D; O(n)</p>
</blockquote>
<h3 id="T-n-x3D-O-n-2"><a href="#T-n-x3D-O-n-2" class="headerlink" title="T(n) &#x3D; O(n^2)"></a>T(n) &#x3D; O(n^2)</h3><figure class="highlight objc"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">- (<span class="type">void</span>)aFunction2:(<span class="type">int</span>)n &#123;</span><br><span class="line">    <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++) &#123; <span class="comment">// n + 1</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> j = <span class="number">0</span>; j &lt; n; j++) &#123; <span class="comment">// n + 1</span></span><br><span class="line">            <span class="built_in">NSLog</span>(<span class="string">@&quot;%zd&quot;</span>,i); <span class="comment">// n</span></span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>执行次数和输入的n值成线性关系。<br>T(n) &#x3D; (n + 1) * (n + 1 + n) &#x3D; 2n^2 + 3n + 1 &#x3D; O(n^2)</p>
</blockquote>
<h3 id="T-n-x3D-O-log-n"><a href="#T-n-x3D-O-log-n" class="headerlink" title="T(n) &#x3D; O(log(n))"></a>T(n) &#x3D; O(log(n))</h3><figure class="highlight objc"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line">- (<span class="type">int</span>)aFunction3:(<span class="type">int</span> *)nums count:(<span class="type">int</span>)count target:(<span class="type">int</span>)target &#123;</span><br><span class="line">    <span class="type">int</span> left = <span class="number">0</span>;</span><br><span class="line">    <span class="type">int</span> right = count - <span class="number">1</span>;</span><br><span class="line">    <span class="type">int</span> mid = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">while</span>(left &lt;= right) &#123;</span><br><span class="line">        mid = (left + right) &gt;&gt; <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span> (nums[mid] &lt; target) &#123;</span><br><span class="line">            left = mid + <span class="number">1</span>;</span><br><span class="line">        &#125; <span class="keyword">else</span> <span class="keyword">if</span> (nums[mid] &gt; target) &#123;</span><br><span class="line">            right = mid - <span class="number">1</span>;</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            <span class="keyword">return</span> mid;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>第1次查找，找到的概率为 1&#x2F;n<br>第2次查找，找到的概率为 2&#x2F;n<br>第3次查找，找到的概率为 4&#x2F;n<br>第m次查找，找到的概率为 2^(m - 1)&#x2F;n<br>假设最多需要查找m次，那么存在：1&#x2F;n + 2&#x2F;n + 4&#x2F;n + … + 2^(m - 1)&#x2F;n &#x3D; 1，可以推导出 m &lt;&#x3D; lg(n + 1)<br>即 T(n) &#x3D; O(log(n))</p>
</blockquote>
<h3 id="T-n-x3D-O-nlog-n"><a href="#T-n-x3D-O-nlog-n" class="headerlink" title="T(n) &#x3D; O(nlog(n))"></a>T(n) &#x3D; O(nlog(n))</h3><figure class="highlight objc"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 快速排序</span></span><br><span class="line">- (<span class="type">void</span>)quickSort:(<span class="type">int</span> *)nums count:(<span class="type">int</span>)count &#123;</span><br><span class="line">    <span class="type">int</span> start = <span class="number">0</span>;</span><br><span class="line">    <span class="type">int</span> end = count - <span class="number">1</span>;</span><br><span class="line">    <span class="type">int</span> value = nums[start];</span><br><span class="line">    </span><br><span class="line">    <span class="keyword">if</span> (count &lt; <span class="number">1</span>) <span class="keyword">return</span>;</span><br><span class="line">    </span><br><span class="line">    <span class="keyword">while</span> (start &lt; end) &#123;</span><br><span class="line">        <span class="comment">//从数组右边往左查找一个小于value的元素</span></span><br><span class="line">        <span class="keyword">while</span> (start &lt; end) &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[end] &lt; value) &#123;</span><br><span class="line">                nums[start] = nums[end];</span><br><span class="line">                start++;</span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">            &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                end--;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">//从数组左边往右查找一个大于value的元素</span></span><br><span class="line">        <span class="keyword">while</span> (start &lt; end) &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[start] &gt; value) &#123;</span><br><span class="line">                nums[end] = nums[start];</span><br><span class="line">                end--;</span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">            &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                start++;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    nums[start] = value;</span><br><span class="line">    </span><br><span class="line">    [<span class="keyword">self</span> quickSort:nums count:start];</span><br><span class="line">    [<span class="keyword">self</span> quickSort:nums + start + <span class="number">1</span> count:count - start - <span class="number">1</span>];</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>第1次递归：T[n] &#x3D; 2T[n&#x2F;2] + n<br>第2次递归：T[n] &#x3D; 2{ 2T[n&#x2F;4] + (n&#x2F;2) }  + n &#x3D; 2^2 T[n&#x2F;(2^2)] + 2n<br>第m次递归：T[n] &#x3D; 2^m T[n&#x2F;(2^m)] + mn<br>假设最多需要m次递归完，那么：T[n&#x2F;(2^m)] &#x3D; T(1)  &#x3D;&#x3D;&gt; m &#x3D; log2(n)<br>得到：T[n] &#x3D; 2^m T[1] + mn &#x3D; 2^(log2(n))T[1] + (log2(n))n &#x3D; nT[1] + (log2(n))n<br>当n趋近于无穷大的时候 T[n] &#x3D; nT[1] + (log2(n))n &#x3D; (log2(n))n &#x3D; O(nlogn)<br>即：T(n) &#x3D; O(nlog(n))   </p>
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